• Use Descartes' Rule of Signs to determine the number of real zeroes of:
  
  f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5.

Descartes' Rule of Signs will not tell you where the polynomial's zeroes are (you'll need to use the Rational Roots Test and synthetic division, or draw a graph, to actually find the roots), but the Rule will tell you how many roots you can expect. First, look at the polynomial as it stands:






f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5






Ignoring the actual values of the coefficients, look at the signs:






f (x) = +x⁵ – x⁴ + 3x³ + 9x² – x + 5






Now note where the signs change from positive to negative or from negative to positive:













Count the number of changes:













There are four sign changes. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial  f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5. However, remember that some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex. Then it may be that certain pairs of roots are not real, and therefore are not graphable as x-intercepts. Because of this possibility, you have to count down by two's. That is, while there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all).




Now look at f (–x):






f (–x) = (–x)⁵ – (–x)⁴ + 3(–x)³ + 9(–x)² – (–x) + 5






= –x⁵ – x⁴ – 3x³ + 9x² + x + 5








Look at the signs:






f (–x) = –x⁵ – x⁴ – 3x³ + 9x² + x + 5






Count the number of sign changes:













There is only one sign change, so there is exactly one negative root. In this case, you don't count down by two's, because you would end up with a negative number.




There are 4, 2, or 0 positive roots, and exactly 1 negative root.

Some texts have you evaluate  f (x) at x = 1 (for the positive roots) and at x = –1 (for the negative roots), so you would get the expressions "1 – 1 + 3 + 9 – 1 + 5" and "–1 – 1 – 3 + 9 + 1 + 5", respectively. You would analyze the signs in the same manner, however.

friend what the iitian from kharagpur said is absolutely correct.

just i m not getting 1 thing.

acc. to me the no of changes in sign by putting +x gives +ve roots and no changes in sign by putting x=-x gives -ve roots if the sum of no of +ve and -ve roots is less than the degree of polynomial then the remaining roots r complex.+ve and -ve roots include all real no rational as well as irrational.